divisibility...

[topspin360]

I used to break down these type of problems through prime factorization. If the overall numerator can have a 9 (ie two 3's), then it is divisible by 9. but it didn't work in this scenario... just looking for a comprehensive thought process on how to go about this problem.

Is x^2 * y^4 an integer divisible by 9 ?

(1) x is an integer divisible by 3

(2) xy is an integer divisible by 9

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.

[GMATGuruNY]

I used to break down these type of problems through prime factorization. If the overall numerator can have a 9 (ie two 3's), then it is divisible by 9. but it didn't work in this scenario... just looking for a comprehensive thought process on how to go about this problem.

Is x^2 * y^4 an integer divisible by 9 ?

(1) x is an integer divisible by 3

(2) xy is an integer divisible by 9

It's important to note how a problem is restricted and how it ISN'T.
There are NO conditions given in the question stem.
Thus, we must consider NONINTEGER values.

If x=18 and y=1/2, both statements are satisfied.
In this case, x²y� = 18²(1/2)�, which is NOT divisible by 9.

If x=18 and y=9, both statements are satisfied.
In this case, x²y� = 18²9�, which IS divisible by 9.

Thus, the two statements combined are INSUFFICIENT.

The correct answer is E.

[Anurag@Gurome]

Deleted.

[eagleeye]

For those of you who are math buffs, here's a mathematical proof of Insufficiency when both statements are taken together.

1. x=3m (where m is an integer)

2. xy=9n (where n is an integer)

Combining the two, (3m)*y=9n => y = 3n/m.

Hence x^2*y^4 = (3m)^2*(3n/m)^4 = 3^6*(n^2/m)^2.

Since n^2/m can be anything ranging from an integer to an even fraction depending upon m and n, The two together are insufficient.

E is correct. Cheers

[Anurag@Gurome]

If x is an integer and and so is xy, y doesn't have to be an integer.

Thanks for pointing it out. Deleted my post.